Two Parallel lines are cut by a transversal x and a transversal y so that x and y intersect at point Q as shown.

Mathematics

1 answer:

saw5 [17]3 years ago

8 0

Answer:

- Angle 'a' is an alternating exterior angle with one angle in the triangle and therefore congruent to one angle in the triangle formed by lines m, x, and y.

- Angles 'b' and 'c' are vertical angles with the other two angles in the triangle and therefore congruent to two other angles in the triangle

Step-by-step explanation:

Vertical angles are angles that are equal to each other but in opposite direction. The angles b and c have vertical angles on the triangle Q while alternate exterior angles are equal angles that lie on different lanes cutting through an axis.

Angles on a plane can be congruent if they are vertical equals or alternating exterior angles.

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Please HELP! I DON'T KNOW HOW CAN I SOLVE THIS

Wewaii [24]

Answer:

Step-by-step explanation:

The Pythagorean Theorem:

$leg^2+leg^2=hypotenuse^2\to a^2+b^2=c^2$

$41.\\\\a=1,\ b=1,\ c=x\\\\x^2=1^2+1^2\\\\x^2=1+1\\\\x^2=2\to x=\sqrt2$

$42.\\a=1, b=2,\ c=x\\\\x^2=1^2+2^2\\\\x^2=1+4\\\\x^2=5\to x=\sqrt5$

$43.\\a=3,\ b=x,\ c=5\\\\3^2+x^2=5^2\\\\9+x^2=25\qquad\text{subtract 9 from both sides}\\\\x^2=16\to x=\sqrt{16}\to x=4$

$44.\\\\a=4,\ b=x,\ c=8\\\\4^2+x^2=8^2\\\\16+x^2=64\qquad\text{subtract 16 from both sides}\\\\x^2=48\to x=\sqrt{48}\\\\x=\sqrt{16\cdot3}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\x=\sqrt{16}\cdot\sqrt3\\\\x=4\sqrt3$